Hydrologic Routing
Reading: Applied Hydrology
Sections 8.1, 8.2, 8.4
Flow Routing
Q
t
Procedure to
determine the
flow hydrograph
at a point on a
watershed from a
known hydrograph
upstream
As the hydrograph
travels, it
attenuates
gets delayed
t
Q
Q
t
2
Why route flows?
Q
Account for changes in flow hydrograph as a flood
wave passes downstream
This helps in
Accounting for storages
Studying the attenuation of flood peaks
3
Types of flow routing
Lumped/hydrologic
Flow is calculated as a function of time
alone at a particular location
Governed by continuity equation and
flow/storage relationship
Distributed/hydraulic
Flow is calculated as a function of space
and time throughout the system
Governed by continuity and momentum
equations
4
Hydrologic Routing
Discharge
I (t )
Inflow
Discharge
Transfer
Function
I (t ) Inflow
Q (t )
Outflow
Q (t ) Outflow
Downstream hydrograph
Upstream
hydrograph
Input, output, and storage are related by continuity
equation:
dS
I (t ) Q (t ) Q and S are
dt
unknown
Storage can be expressed as a function of I(t) or Q(t) or
both
S f (I ,
dI
dQ
, , Q,
, )
dt
dt
For a linear reservoir, S=kQ
Lumped flow routing
Three types
1. Level pool method (Modified Puls)
Storage is nonlinear function of Q
2. Muskingum method
Storage is linear function of I and Q
3. Series of reservoir models
Storage is linear function of Q and its
time derivatives
6
S and Q relationships
Level pool routing
Procedure for calculating outflow
hydrograph Q(t) from a reservoir with
horizontal water surface, given its
inflow hydrograph I(t) and storageoutflow relationship
Level pool methodology
Discharge
dS
I (t ) Q(t )
dt
Inflow
I j 1
Outflow
S j 1
( j 1) t
( j 1) t
Sj
jt
j t
dS
Ij
Q j 1
Qj
S j 1 S j
t
jt
( j 1) t
Time
t
2 S j 1
Storage
Idt
Qdt
I j 1 I j
2
Q j 1 Q j
Q j 1 I j 1 I j
Unknown
Sj
9
Time
2S j
Known
Need a function relating
S j 1
2S
Q, and Q
t
Storage-outflow function
Qj
Level pool methodology
Given
Inflow hydrograph
Q and H relationship
Steps
1. Develop Q versus Q+ 2S/t
relationship using Q/H relationship
2 S j 1
2S j
2. Compute Q+ 2S/t using Q j 1 I j 1 I j Q j
t
t
3. Use the relationship developed in step
1 to get Q
10
Ex. 8.2.1
Given I(t)
Given
Q/H
Elevation H Discharge Q
(ft)
(cfs)
0
0
0.5
3
1
8
1.5
17
2
30
2.5
43
3
60
3.5
78
4
97
4.5
117
5
137
5.5
156
6
173
6.5
190
7
205
7.5
218
8
231
8.5
242
9
253
9.5
264
10
275
Area of the reservoir = 1 acre, and outlet
diameter = 5ft
11
Ex. 8.2.1 Step 1
evelop Q versus Q+ 2S/t relationship using Q/H relationship
Elevation H Discharge Q Storage S 2S/ t + Q
3
(ft )
(ft)
(cfs)
(cfs)
0
0
0
0
0.5
3
21780
75.6
1
8
43560
153.2
1.5
17
65340
234.8
2
30
87120
320.4
2.5
43
108900
406
3
60
130680
495.6
3.5
78
152460
586.2
4
97
174240
677.8
4.5
117
196020
770.4
5
137
217800
863
5.5
156
239580
954.6
6
173
261360
1044.2
6.5
190
283140
1133.8
7
205
304920
1221.4
7.5
218
326700
1307
8
231
348480
1392.6
8.5
242
370260
1476.2
9
253
392040
1559.8
9.5
264
413820
1643.4
10
275
435600
1727
S Area Height 43560 0.5 21,780 ft 3
2S
2 21780
Q
3 75.6 cfs
t
10 60
12
Step 2
Compute Q+ 2S/t using
2 S j 1
t
Q j 1 I j 1 I j
2S j
t
Qj
At time interval =1 (j=1), I1 = 0, and therefore Q1 = 0 as the reservoir
is empty
Write the continuity equation for the first time step,
which can be used to compute Q2
2S 2
2S1
Q2 I 2 I1
Q1
2S 2
2 S1
Q2 I 2 I1
Q1 0 60 60
13
Step 3
Use the relationship between 2S/t + Q
versus Q to compute Q
2S 2
Q2 60
e the Table/graph created in Step 1 to compute Q
What is the value of Q if 2S/t + Q
= 60 ?
(3 0)
Q 0
(60 0) 2.4 cfs
(76 0)
So Q2 is 2.4
cfs
Repeat steps 2 and 3 for j=2, 3, 4 to
compute Q3, Q4, Q5..
14
Elevation H Discharge Q Storage S 2S/ t + Q
3
(ft )
(ft)
(cfs)
(cfs)
0
0
0
0
0.5
3
21780
75.6
1
8
43560
153.2
1.5
17
65340
234.8
2
30
87120
320.4
2.5
43
108900
406
3
60
130680
495.6
3.5
78
152460
586.2
4
97
174240
677.8
4.5
117
196020
770.4
5
137
217800
863
5.5
156
239580
954.6
6
173
261360
1044.2
6.5
190
283140
1133.8
7
205
304920
1221.4
7.5
218
326700
1307
8
231
348480
1392.6
8.5
242
370260
1476.2
9
253
392040
1559.8
9.5
264
413820
1643.4
10
275
435600
1727
Ex. 8.2.1 results
2S j
t
Qj
2 S j 1
2S j
t
Q j 2Q j
15
Q j 1 I j 1 I j
2S j
t
Qj
Ex. 8.2.1 results
12.0
Storage (acre-ft)
10.0
Outflow
hydrograp
h
8.0
6.0
4.0
2.0
0.0
0
20
40
60
80
100
120
140
160
180
200
220
Time (minutes)
400
350
Inflow
Discharge (cfs)
300
Peak outflow intersects with the
receding limb of the inflow
hydrograph
250
200
150
Outflow
100
50
0
0
20
40
60
80
100
120
TIme (minutes)
16
140
160
180
200
220
Q/H relationships
http://www.wsi.nrcs.usda.gov/products/W2Q/H&H/Tools_Models/Sites
.html
17 an NRCS Reservoir
Program for Routing Flow through
Hydrologic Reservoir Routing
Assuming that cross sectional area of flood flow is directly
proportional to discharge.
Hydrologic river routing (Muskingum Method)
Wedge storage in reach
S Prism KQ
S Wedge KX ( I Q )
Advancing
Flood
Wave
I>Q
K = travel time of peak through the reach
X = weight on inflow versus outflow (0 X
0.5)
X = 0 Reservoir, storage depends on
outflow, no wedge
X = 0.0 - 0.3 Natural stream
S KQ KX ( I Q)
S K [ XI (1 X )Q]
I Q
Q
Receding
Flood
Wave
Q>I
QI
I
Muskingum Method (Cont.)
S K [ XI (1 X )Q]
S j 1 S j K {[ XI j 1 (1 X )Q j 1 ] [ XI j (1 X )Q j ]}
Recall:
S j 1 S j
I j 1 I j
2
Q j 1 Q j
2
Combine:
Q j 1 C1 I j 1 C 2 I j C3Q j
t 2 KX
2 K (1 X ) t
t 2 KX
C2
2 K (1 X ) t
2 K (1 X ) t
C3
2 K (1 X ) t
C1
If I(t), K and X are known, Q(t) can be calculated using
20
above equations
Muskingum - Example
Given:
Inflow hydrograph
K = 2.3 hr, X = 0.15, t =
1 hour, Initial Q = 85 cfs
Find:
Outflow hydrograph using
Muskingum routing
method
t 2 KX
1 2 * 2.3 * 0.15
0.0631
2 K (1 X ) t 2 * 2.3(1 0.15) 1
t 2 KX
1 2 * 2.3 * 0.15
C2
0.3442
2 K (1 X ) t 2 * 2.3(1 0.15) 1
2 K (1 X ) t 2 * 2.3 * (1 0.15) 1
C3
0.5927
2 K (1 X ) t
2 * 2.3(1 0.15) 1
C1
21
Muskingum Example (Cont.)
Q j 1 C1I j 1 C 2 I j C3Q j
C1 = 0.0631, C2 = 0.3442, C3
= 0.5927
800
700
Discharge (cfs)
600
500
400
300
200
100
0
1
10 11 12 13 14 15 16 17 18 19 20
Time (hr)
22
